Let $g(x)=\dfrac{x-5}{x^2+1}$. $g'(x)=$
Answer: $g$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = g ′ ( x ) = d d x ( x − 5 x 2 + 1 ) = ( x 2 + 1 ) d d x ( x − 5 ) − ( x − 5 ) d d x ( x 2 + 1 ) ( x 2 + 1 ) 2 = ( x 2 + 1 ) ( 1 ) − ( x − 5 ) ( 2 x ) ( x 2 + 1 ) 2 = x 2 + 1 − 2 x 2 + 10 x ( x 2 + 1 ) 2 = 1 + 10 x − x 2 ( x 2 + 1 ) 2 The quotient rule Differentiate ( x − 5 ) & ( x 2 + 1 ) Expand \begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x-5}{x^2+1}\right) \\\\ &=\dfrac{(x^2+1)\dfrac{d}{dx}(x-5)-(x-5)\dfrac{d}{dx}(x^2+1)}{(x^2+1)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x^2+1)(1)-(x-5)(2x)}{(x^2+1)^2}&&\gray{\text{Differentiate }(x-5)\text{ & }(x^2+1)} \\\\ &=\dfrac{x^2+1-2x^2+10x}{(x^2+1)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{1+10x-x^2}{(x^2+1)^2} \end{aligned} In conclusion, $g'(x)=\dfrac{1+10x-x^2}{(x^2+1)^2}$, or any other equivalent form.